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Problem 26 Hard Difficulty

A $25.0-\mathrm{pm}$ x-ray photon scatters off a free electron at $A$ (Fig. $\mathrm{P} 27.26 ),$ producing a photon of wavelength $\lambda^{\prime}$ traveling at an angle $\theta=40.0^{\circ}$ relative to the first photon's direction. This second photon scatters off another free electron at $B,$ producing a photon with wavelength $\lambda^{\prime \prime}$ and moving in a direction directly opposite the first photon. Determine the wavelengths (a) $\lambda^{\prime}$ and ( b ) $\lambda^{\prime \prime}$

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg

Video Transcript

Yes. This is kind of interesting question. Um, so we have an X ray. That's 225 PICO meters. So let's go ahead and write that down. Um, 25 times, 10 to the minus 12 meters and, um, free electron A A. So I guess there's, like, a nice figure. Oh, yeah. And then and then it gets scattered, too. Point b. And then? And then it gets scattered. Um, okay, then I get scattered again, and then it's you got this Lambda double prime. And so we want to get land. Ah, Prime and Lambda Double prime. So we want to use this. Um, let's see, we want to use the Compton effect. So that tells you that, um, alto and General Delta Lambda is equal to H Over. Um see, Time's one minus co sign If I and then let's go ahead. I'm just gonna This looks like a mass. My rewrite this so co sign by and then the coast signifies the angle regular deviation from going, being going straight forward. So let's get the adult of fire exploring the Delta Lambda for the first process. So number one your Delta Lambda is gonna be like an H over M. C. And of course, Abbas, the mass of the electron. Um and then we can see that's deflected by 40 degrees. So it's gonna be you, um, 40 degrees here. So let's go ahead and calculate that. So age divided by M. C. One line is co sign of 40. And then I got bread. Delta Lambda is 5.7. Sorry. 5.68 times 10 to the minus 13. So that's the change in and the way blank. And so then therefore, Lambda Prime is gonna be the change plus the original. So we want to do I'll call this Lambda Ice just lambda still total. And, uh, I'll call this Delta Lambda one just to distinguish, cause we're gonna have another change of Lambda. So let's go ahead and add these numbers up. So take this plus 25 times 10 to the minus 12 and we gotta get 2.56 times 10 to the minus 11 um, mirrors. And now we have this same to prime. So let's go get our land a double prime so we can see that, um, cm probably to draw this picture out. Okay, so we need to get that the angle that this changes through because we have Oh, no. Let me rewind. Um oh, yeah, because we actually Oh, yeah. We actually know how much the angle changes. And so, um so let's go ahead and draw it out and get the angle that it's scattered through. So the incoming one is going like this, and then this is this is the limit of prime, and in the outgoing one is going like this, so this one needs to rotate by this full angle. So let's go ahead and her figure out what that ankle is. Um, so just from this picture, we know that, um, that angle See, actually, I'm gonna try to draw a better picture. Thio clearly demonstrate everything that's going on. Um, so you have the incident full time coming in, gets scattered this way, and then we know that this is 40 degrees, and then it gets scattered this way. So it's angle's changing quite a bit. It's going kind of don and to the right and then sweeping up. So, um, I'm just gonna draw this down here and say that we we need to prime this angle. So this one's 40. This is 140 because these have to add up to 1 80 So I would say that this angle is 140 because this both of these signs of peril. So these two are parallel and this is the same lie. So of course it's parallel to itself. So then, in that case, we're gonna get to find the shift. We want to use 140 degrees so we can get the Delta Phi to their sorry dot Talking landed too. This h over M C. Times one minus CO sign of 40. And so let's go ahead and calculate that to get this shift, so should be a bigger Shipton. And it s so that's 4.28 times 10 to the minus 12 meters. And what we want to do is take that in. Added to this to get Lambda double prime. So then we can say Lambda double prime is gonna be the end of prime plus Delta Lambda +01 So four plus 4.28 times 10 of them. I was 12 so that I get 8.56 times 10 to the minus 12 meters, right

University of Washington
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg